cos(2π)/(15)cos(4π)/(15)cos(8π)/(15)cos(16π)/(15) = (1)/(16).

NCERT Exemplar · Q71

\cos\dfrac{2\pi}{15}\,\cos\dfrac{4\pi}{15}\,\cos\dfrac{8\pi}{15}\,\cos\dfrac{16\pi}{15} = \dfrac{1}{16}

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Concept understanding — Trigonometric Functions in Quadrants

Trigonometric Functions in Quadrants

Imagine standing at the centre of a circle, facing east. If you turn by some angle, you end up pointing in a certain direction. That direction has both a horizontal component (east-west) and a vertical component (north-south). Trigonometric functions are just a way to describe those components — and whether they are positive or negative depends entirely on which quadrant you're facing.

The Four Quadrants

The coordinate plane is split into four quadrants, numbered anticlockwise starting from the top-right:

Quadrant I (0° to 90°): x > 0, y > 0
Quadrant II (90° to 180°): x < 0, y > 0
Quadrant III (180° to 270°): x < 0, y < 0
Quadrant IV (270° to 360°): x > 0, y < 0

Now, recall the definitions on the unit circle (radius = 1):

$\cos \theta$ = x-coordinate of the point on the circle
$\sin \theta$ = y-coordinate of that point
$\tan \theta = \frac{\sin \theta}{\cos \theta}$

So the sign of $\cos \theta$ follows the sign of x, and the sign of $\sin \theta$ follows the sign of y. That's all there is to it.

The Sign Pattern

Quadrant	$\sin \theta$	$\cos \theta$	$\tan \theta$
I (0–90)	+	+	+
II (90–180)	+	–	–
III (180–270)	–	–	+
IV (270–360)	–	+	–

Tip

The mnemonic "All Students Take Coffee" helps you remember which functions are positive in each quadrant, starting from QI and going anticlockwise: All (all positive), Sine (sin positive), Tan (tan positive), Cos (cos positive).

Why This Matters

Suppose you're solving $\sin \theta = \frac{1}{2}$ . The calculator gives you $\theta = 30^\circ$ , but that's only one solution. Because sine is positive in both QI and QII, there's a second angle: $180^\circ - 30^\circ = 150^\circ$ . If you forget the quadrant rule, you lose half the answers.

Similarly, if $\cos \theta = -\frac{\sqrt{3}}{2}$ , cosine is negative in QII and QIII. So the solutions are $150^\circ$ and $210^\circ$ (plus full rotations). …

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Using the telescoping identity $\sin(2^n\theta) = 2^n\sin\theta\cos\theta\cos2\theta\cdots\cos(2^{n-1}\theta)$ with $\theta=\dfrac{2\pi}{15}$ , the product collapses to exactly $\dfrac{1}{16}$ — the given statement is true.

Setting up the telescoping pattern

Let $\theta = \dfrac{2\pi}{15}$ . Then $2\theta = \dfrac{4\pi}{15}$ , $4\theta = \dfrac{8\pi}{15}$ , and $8\theta = \dfrac{16\pi}{15}$ — exactly the four angles in the product. So we need

P = \cos\theta \cdot \cos2\theta \cdot \cos4\theta \cdot \cos8\theta.

Step 1 — Multiply and divide by $\sin\theta$ .

Since $\theta = \dfrac{2\pi}{15}$ is not a multiple of $\pi$ , $\sin\theta \neq 0$ , so this is valid:

P = \frac{\sin\theta\cos\theta\cos2\theta\cos4\theta\cos8\theta}{\sin\theta}.

Step 2 — Apply $\sin 2A = 2\sin A\cos A$ repeatedly, absorbing one cosine at a time.

\sin\theta\cos\theta = \tfrac{1}{2}\sin2\theta \implies \text{numerator} = \tfrac{1}{2}\sin2\theta\cos2\theta\cos4\theta\cos8\theta

\sin2\theta\cos2\theta = \tfrac{1}{2}\sin4\theta \implies \text{numerator} = \tfrac{1}{4}\sin4\theta\cos4\theta\cos8\theta

$\sin4\theta\cos4\theta = \tfrac{1}{2}\sin8\theta \implies \text{numerator} = \tfrac{1}{8}\sin8\theta\cos8\theta$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04184 marksMCQ
$\sin\theta=\dfrac{3}{5}$
›Reveal solutionSolution

With $\cos\theta=-\tfrac45$ , $\tan\theta=\tfrac{3/5}{-4/5}=-\tfrac34$ .

From $\sin\theta=\tfrac35$ , $\cos^{2}\theta=1-\tfrac{9}{25}=\tfrac{16}{25}$ , so $\cos\theta=\pm\tfrac45$ . Given $\cos\theta<0$ , $\cos\theta=-\tfrac45$ . …
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KEAM 2026Set eng-2026-04194 marksMCQ
$\sin\theta \cos\theta > 0$
›Reveal solutionSolution
The product $\sin\theta\cos\theta>0$ means $\sin\theta$ and $\cos\theta$ have the same sign: quadrant I or III.

Sign check by quadrant:

I: $\sin>0,\cos>0\Rightarrow$ product $>0$ . ✓

II: $\sin>0,\cos<0\Rightarrow$ product $<0$ .

III: $\sin<0,\cos<0\Rightarrow$ product $>0$ . ✓ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$3\cot(-405^\circ)\tan 315^\circ - 5\cot 495^\circ \tan(-585^\circ)$
›Reveal solutionSolution
Reduce each angle: $\cot(-405^\circ)=-1$ , $\tan315^\circ=-1$ , $\cot495^\circ=-1$ , $\tan(-585^\circ)=-1$ . Substituting gives $3(1)-5(1)=-2$ .

Evaluate each term using periodicity and sign rules:

$\cot(-405^\circ)=-\cot405^\circ=-\cot45^\circ=-1$ .

$\tan315^\circ=\tan(315^\circ-360^\circ)=\tan(-45^\circ)=-1$ .

$\cot495^\circ=\cot(495^\circ-360^\circ)=\cot135^\circ=-1$ . …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\sin \alpha = \dfrac{12}{13}$
›Reveal solutionSolution
With sin alpha = 12/13 positive in the given range, alpha is in Q2, so cos = -5/13 and tan alpha = -12/5.

Concept and Intuition

On (pi/2, 3pi/2), sine is positive only in the second quadrant (pi/2, pi), where cosine is negative. That fixes the sign of the tangent.

Step-by-Step Solution

sin alpha = 12/13 > 0 with pi/2 < alpha < 3pi/2 forces alpha in (pi/2, pi).

cos alpha = -sqrt(1 - 144/169) = -5/13. …
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KEAM 2025Set eng-2025-04264 marksMCQ
$\sin\theta = \frac{1}{5}$
›Reveal solutionSolution

$\sin\theta=\frac15$ , second quadrant $\Rightarrow\cos\theta=-\frac{2\sqrt6}{5}$ ; reciprocal gives $\sec\theta=-\frac{5}{2\sqrt6}$ . …

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KEAM 2025Set eng-2025-04274 marksMCQ
$\tan(315^\circ)\cot(-405^\circ) =$
›Reveal solutionSolution

$\tan315^\circ=-1$ and $\cot(-405^\circ)=-1$ , giving a product of $1$ .

Evaluate each factor.

$\tan315^\circ=\tan(360^\circ-45^\circ)=-\tan45^\circ=-1.$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$\dfrac{1}{1-\tan x}=\dfrac{3+\sqrt{3}}{2}$
›Reveal solutionSolution

Invert to solve for $\tan x$ . …

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KEAM 2024Set eng-2024-06074 marksMCQ
$\tan\left(\cos^{-1}\left(\dfrac{-24}{25}\right)\right)$
›Reveal solutionSolution

Let $\theta=\cos^{-1}(-24/25)$ , so $\theta\in[0,\pi]$ lies in the second quadrant. There $\sin\theta>0$ , so $\sin\theta=+\sqrt{1-(24/25)^2}=7/25$ , and $\tan\theta=\dfrac{7/25}{-24/25}=-\dfrac{7}{24}$ .

Set $\theta=\cos^{-1}\left(\dfrac{-24}{25}\right)$ . The range of $\cos^{-1}$ is $[0,\pi]$ , and since the cosine is negative, $\theta$ is in $(\pi/2,\pi)$ — the second quadrant, where $\sin\theta>0$ . …
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KEAM 2024Set eng-2024-06084 marksMCQ
$\cos x-\sin x=0$
›Reveal solutionSolution

$\tan x=1$ restricted to $[0,\pi]$ gives only $x=\tfrac{\pi}{4}$ .

From $\cos x-\sin x=0$ we get $\tan x=1$ . The general solution is $x=\dfrac{\pi}{4}+n\pi$ . …
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KEAM 2024Set eng-2024-06084 marksMCQ
$2\sin\left(\dfrac{\pi}{3}-2x\right)-1=0$
›Reveal solutionSolution
Solve $\sin(\tfrac{\pi}{3}-2x)=\tfrac12$ ; only $x=\tfrac{\pi}{12}$ lies in $(0,\tfrac{\pi}{2})$ .

From $2\sin\!\left(\dfrac{\pi}{3}-2x\right)-1=0$ we get $\sin\!\left(\dfrac{\pi}{3}-2x\right)=\dfrac12$ .

So $\dfrac{\pi}{3}-2x=\dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$ .

$\dfrac{\pi}{3}-2x=\dfrac{\pi}{6}\Rightarrow 2x=\dfrac{\pi}{6}\Rightarrow x=\dfrac{\pi}{12}$ . …
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KEAM 2024Set eng-2024-06084 marksMCQ
$\sin x = \frac{3}{5}$
›Reveal solutionSolution

Take the principal (first-quadrant) value $\cos x=\tfrac45$ and evaluate.

Given $\sin x=\dfrac35$ , so $\cos x=\sqrt{1-\tfrac{9}{25}}=\dfrac45$ (taking the positive value). …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$\cos\theta = \frac{5}{11}$
›Reveal solutionSolution
$\sin\theta=-\frac{4\sqrt6}{11}$ .

Concept and Intuition

Use $\sin^2\theta+\cos^2\theta=1$ and fix the sign from the quadrant implied by the sign of $\tan\theta$ .

Step-by-Step Solution

$\cos\theta=\frac{5}{11}>0$ ; $\tan\theta<0$ means $\sin\theta$ and $\cos\theta$ have opposite signs, so $\sin\theta<0$ (quadrant IV).

$\sin^2\theta=1-\frac{25}{121}=\frac{96}{121}$ .

$\sin\theta=-\sqrt{\frac{96}{121}}=-\frac{\sqrt{96}}{11}$ . …
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←Previous questionQ70 — sin 10^° is greater than cos 10^°.→Next questionQ72 — One value of θ which satisfies the equation sin⁴θ - 2sin²θ - 1 lies be…

Q.cos⁡2π15 cos⁡4π15 cos⁡8π15 cos⁡16π15=116\cos\dfrac{2\pi}{15}\,\cos\dfrac{4\pi}{15}\,\cos\dfrac{8\pi}{15}\,\cos\dfrac{16\pi}{15} = \dfrac{1}{16}cos152π​cos154π​cos158π​cos1516π​=161​.

Setting up the telescoping pattern

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Q. $\cos\dfrac{2\pi}{15}\,\cos\dfrac{4\pi}{15}\,\cos\dfrac{8\pi}{15}\,\cos\dfrac{16\pi}{15} = \dfrac{1}{16}$ .

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